mehr1methanol
- 10
- 0
Example where X is not in the Lebesgue linear space.
I'm trying to find an example where [itex]\lim_{n \to +\infty} P(|X|>n) = 0[/itex] but [itex]X \notin L[/itex] where [itex]L[/itex] is the Lebesgue linear space.
Relevant equations:
[itex]X[/itex] is a random variabel, [itex]P[/itex] is probability. [itex]I[/itex] is indicator function.
The attempt at a solution
[itex]∫|X|I(|X|>n)dp + ∫|X|I(|X|≤n)dp = ∫|X|dp[/itex]
[itex]∫nI(|X|>n)dp + ∫|X|I(|X|)dp ≤ ∫|X|dp[/itex]
Suppose [itex]∫I(|X|>n)dp = 1/(n ln n)[/itex]
Clearly the hypothesis is satisfied because [itex]\lim_{n \to +\infty} P(|X|>n) = \lim_{n \to +\infty} ∫I(|X|>n)dp = \lim_{n \to +\infty} 1/( ln n) = 0[/itex]
But I'm not sure how to conclude [itex]∫|X|dp = ∞[/itex]
Homework Statement
I'm trying to find an example where [itex]\lim_{n \to +\infty} P(|X|>n) = 0[/itex] but [itex]X \notin L[/itex] where [itex]L[/itex] is the Lebesgue linear space.
Relevant equations:
[itex]X[/itex] is a random variabel, [itex]P[/itex] is probability. [itex]I[/itex] is indicator function.
The attempt at a solution
[itex]∫|X|I(|X|>n)dp + ∫|X|I(|X|≤n)dp = ∫|X|dp[/itex]
[itex]∫nI(|X|>n)dp + ∫|X|I(|X|)dp ≤ ∫|X|dp[/itex]
Suppose [itex]∫I(|X|>n)dp = 1/(n ln n)[/itex]
Clearly the hypothesis is satisfied because [itex]\lim_{n \to +\infty} P(|X|>n) = \lim_{n \to +\infty} ∫I(|X|>n)dp = \lim_{n \to +\infty} 1/( ln n) = 0[/itex]
But I'm not sure how to conclude [itex]∫|X|dp = ∞[/itex]
Last edited: